正项数列{an}满足
(1)求数列{an}的通项公式an; (2)令bn=
|
(1)由正项数列{an}满足:
-(2n-1)an-2n=0,a 2n
可得(an-2n)(an+1)=0
所以an=2n.
(2)因为an=2n,bn=
,1 (n+1)an
所以bn=1 (n+1)an
=1 2n(n+1)
=
(1 2
-1 n
),1 n+1
Tn=
(1-1 2
+1 2
-1 2
+…+1 3
-1 n
)1 n+1
=
(1-1 2
)1 n+1
=
.n 2n+2
数列{bn}的前n项和Tn为
.n 2n+2